Integrand size = 120, antiderivative size = 27 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=x+x^2 \log ^2(x)+\frac {25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )}{(-1+\log (4))^2} \] Output:
625*ln(ln(1/2*x))^4/(10*ln(2)-5)^2+x+x^2*ln(x)^2
Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=x+x^2 \log ^2(x)+\frac {25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )}{(-1+\log (4))^2} \] Input:
Integrate[((x - 2*x*Log[4] + x*Log[4]^2)*Log[x/2] + (2*x^2 - 4*x^2*Log[4] + 2*x^2*Log[4]^2)*Log[x/2]*Log[x] + (2*x^2 - 4*x^2*Log[4] + 2*x^2*Log[4]^2 )*Log[x/2]*Log[x]^2 + 100*Log[Log[x/2]]^3)/((x - 2*x*Log[4] + x*Log[4]^2)* Log[x/2]),x]
Output:
x + x^2*Log[x]^2 + (25*Log[Log[x/2]]^4)/(-1 + Log[4])^2
Time = 0.48 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6, 6, 27, 7239, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (2 x^2+2 x^2 \log ^2(4)-4 x^2 \log (4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+\left (2 x^2+2 x^2 \log ^2(4)-4 x^2 \log (4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )+\left (x+x \log ^2(4)-2 x \log (4)\right ) \log \left (\frac {x}{2}\right )}{\left (x+x \log ^2(4)-2 x \log (4)\right ) \log \left (\frac {x}{2}\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (2 x^2+2 x^2 \log ^2(4)-4 x^2 \log (4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+\left (2 x^2+2 x^2 \log ^2(4)-4 x^2 \log (4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )+\left (x+x \log ^2(4)-2 x \log (4)\right ) \log \left (\frac {x}{2}\right )}{\left (x \log ^2(4)+x (1-2 \log (4))\right ) \log \left (\frac {x}{2}\right )}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (2 x^2+2 x^2 \log ^2(4)-4 x^2 \log (4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+\left (2 x^2+2 x^2 \log ^2(4)-4 x^2 \log (4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )+\left (x+x \log ^2(4)-2 x \log (4)\right ) \log \left (\frac {x}{2}\right )}{x \left (1+\log ^2(4)-2 \log (4)\right ) \log \left (\frac {x}{2}\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )+2 \left (\log ^2(4) x^2-2 \log (4) x^2+x^2\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+x (1-\log (4))^2 \log \left (\frac {x}{2}\right )+2 \left (\log ^2(4) x^2-2 \log (4) x^2+x^2\right ) \log \left (\frac {x}{2}\right ) \log (x)}{x \log \left (\frac {x}{2}\right )}dx}{(1-\log (4))^2}\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \frac {\int \left (\frac {100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{x \log \left (\frac {x}{2}\right )}+(-1+\log (4))^2 \left (2 x \log ^2(x)+2 x \log (x)+1\right )\right )dx}{(1-\log (4))^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^2 (1-\log (4))^2 \log ^2(x)+25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )+x (1-\log (4))^2}{(1-\log (4))^2}\) |
Input:
Int[((x - 2*x*Log[4] + x*Log[4]^2)*Log[x/2] + (2*x^2 - 4*x^2*Log[4] + 2*x^ 2*Log[4]^2)*Log[x/2]*Log[x] + (2*x^2 - 4*x^2*Log[4] + 2*x^2*Log[4]^2)*Log[ x/2]*Log[x]^2 + 100*Log[Log[x/2]]^3)/((x - 2*x*Log[4] + x*Log[4]^2)*Log[x/ 2]),x]
Output:
(x*(1 - Log[4])^2 + x^2*(1 - Log[4])^2*Log[x]^2 + 25*Log[Log[x/2]]^4)/(1 - Log[4])^2
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26
| method | result | size |
| parts | \(x +x^{2} \ln \left (x \right )^{2}+\frac {25 \ln \left (\ln \left (\frac {x}{2}\right )\right )^{4}}{4 \ln \left (2\right )^{2}-4 \ln \left (2\right )+1}\) | \(34\) |
| default | \(x +x^{2} \ln \left (x \right )^{2}+\frac {25 \ln \left (-\ln \left (2\right )+\ln \left (x \right )\right )^{4}}{4 \ln \left (2\right )^{2}-4 \ln \left (2\right )+1}\) | \(37\) |
| risch | \(x +x^{2} \ln \left (x \right )^{2}+\frac {25 \ln \left (-\ln \left (2\right )+\ln \left (x \right )\right )^{4}}{4 \ln \left (2\right )^{2}-4 \ln \left (2\right )+1}\) | \(37\) |
| parallelrisch | \(\frac {x -4 x \ln \left (2\right )+4 x \ln \left (2\right )^{2}+x^{2} \ln \left (x \right )^{2}+25 \ln \left (\ln \left (\frac {x}{2}\right )\right )^{4}+4 x^{2} \ln \left (2\right )^{2} \ln \left (x \right )^{2}-4 \ln \left (2\right ) \ln \left (x \right )^{2} x^{2}}{4 \ln \left (2\right )^{2}-4 \ln \left (2\right )+1}\) | \(71\) |
Input:
int((100*ln(ln(1/2*x))^3+(8*x^2*ln(2)^2-8*x^2*ln(2)+2*x^2)*ln(1/2*x)*ln(x) ^2+(8*x^2*ln(2)^2-8*x^2*ln(2)+2*x^2)*ln(1/2*x)*ln(x)+(4*x*ln(2)^2-4*x*ln(2 )+x)*ln(1/2*x))/(4*x*ln(2)^2-4*x*ln(2)+x)/ln(1/2*x),x,method=_RETURNVERBOS E)
Output:
x+x^2*ln(x)^2+25/(4*ln(2)^2-4*ln(2)+1)*ln(ln(1/2*x))^4
Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (27) = 54\).
Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 4.41 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=\frac {4 \, x^{2} \log \left (2\right )^{4} - 4 \, x^{2} \log \left (2\right )^{3} + 25 \, \log \left (\log \left (\frac {1}{2} \, x\right )\right )^{4} + {\left (x^{2} + 4 \, x\right )} \log \left (2\right )^{2} + {\left (4 \, x^{2} \log \left (2\right )^{2} - 4 \, x^{2} \log \left (2\right ) + x^{2}\right )} \log \left (\frac {1}{2} \, x\right )^{2} - 4 \, x \log \left (2\right ) + 2 \, {\left (4 \, x^{2} \log \left (2\right )^{3} - 4 \, x^{2} \log \left (2\right )^{2} + x^{2} \log \left (2\right )\right )} \log \left (\frac {1}{2} \, x\right ) + x}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} \] Input:
integrate((100*log(log(1/2*x))^3+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1 /2*x)*log(x)^2+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1/2*x)*log(x)+(4*x* log(2)^2-4*x*log(2)+x)*log(1/2*x))/(4*x*log(2)^2-4*x*log(2)+x)/log(1/2*x), x, algorithm="fricas")
Output:
(4*x^2*log(2)^4 - 4*x^2*log(2)^3 + 25*log(log(1/2*x))^4 + (x^2 + 4*x)*log( 2)^2 + (4*x^2*log(2)^2 - 4*x^2*log(2) + x^2)*log(1/2*x)^2 - 4*x*log(2) + 2 *(4*x^2*log(2)^3 - 4*x^2*log(2)^2 + x^2*log(2))*log(1/2*x) + x)/(4*log(2)^ 2 - 4*log(2) + 1)
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=x^{2} \log {\left (x \right )}^{2} + x + \frac {25 \log {\left (\log {\left (x \right )} - \log {\left (2 \right )} \right )}^{4}}{- 4 \log {\left (2 \right )} + 1 + 4 \log {\left (2 \right )}^{2}} \] Input:
integrate((100*ln(ln(1/2*x))**3+(8*x**2*ln(2)**2-8*x**2*ln(2)+2*x**2)*ln(1 /2*x)*ln(x)**2+(8*x**2*ln(2)**2-8*x**2*ln(2)+2*x**2)*ln(1/2*x)*ln(x)+(4*x* ln(2)**2-4*x*ln(2)+x)*ln(1/2*x))/(4*x*ln(2)**2-4*x*ln(2)+x)/ln(1/2*x),x)
Output:
x**2*log(x)**2 + x + 25*log(log(x) - log(2))**4/(-4*log(2) + 1 + 4*log(2)* *2)
Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (27) = 54\).
Time = 0.24 (sec) , antiderivative size = 354, normalized size of antiderivative = 13.11 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=\frac {75 \, \log \left (-\log \left (2\right ) + \log \left (x\right )\right )^{4}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} - \frac {150 \, \log \left (-\log \left (2\right ) + \log \left (x\right )\right )^{2} \log \left (\log \left (\frac {1}{2} \, x\right )\right )^{2}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {100 \, \log \left (-\log \left (2\right ) + \log \left (x\right )\right ) \log \left (\log \left (\frac {1}{2} \, x\right )\right )^{3}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {2 \, {\left (2 \, x^{2} \log \left (x\right )^{2} - 2 \, x^{2} \log \left (x\right ) + x^{2}\right )} \log \left (2\right )^{2}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {2 \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} \log \left (2\right )^{2}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {4 \, x \log \left (2\right )^{2}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} - \frac {2 \, {\left (2 \, x^{2} \log \left (x\right )^{2} - 2 \, x^{2} \log \left (x\right ) + x^{2}\right )} \log \left (2\right )}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} - \frac {2 \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} \log \left (2\right )}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} - \frac {4 \, x \log \left (2\right )}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {2 \, x^{2} \log \left (x\right )^{2} - 2 \, x^{2} \log \left (x\right ) + x^{2}}{2 \, {\left (4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1\right )}} + \frac {2 \, x^{2} \log \left (x\right ) - x^{2}}{2 \, {\left (4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1\right )}} + \frac {x}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} \] Input:
integrate((100*log(log(1/2*x))^3+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1 /2*x)*log(x)^2+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1/2*x)*log(x)+(4*x* log(2)^2-4*x*log(2)+x)*log(1/2*x))/(4*x*log(2)^2-4*x*log(2)+x)/log(1/2*x), x, algorithm="maxima")
Output:
75*log(-log(2) + log(x))^4/(4*log(2)^2 - 4*log(2) + 1) - 150*log(-log(2) + log(x))^2*log(log(1/2*x))^2/(4*log(2)^2 - 4*log(2) + 1) + 100*log(-log(2) + log(x))*log(log(1/2*x))^3/(4*log(2)^2 - 4*log(2) + 1) + 2*(2*x^2*log(x) ^2 - 2*x^2*log(x) + x^2)*log(2)^2/(4*log(2)^2 - 4*log(2) + 1) + 2*(2*x^2*l og(x) - x^2)*log(2)^2/(4*log(2)^2 - 4*log(2) + 1) + 4*x*log(2)^2/(4*log(2) ^2 - 4*log(2) + 1) - 2*(2*x^2*log(x)^2 - 2*x^2*log(x) + x^2)*log(2)/(4*log (2)^2 - 4*log(2) + 1) - 2*(2*x^2*log(x) - x^2)*log(2)/(4*log(2)^2 - 4*log( 2) + 1) - 4*x*log(2)/(4*log(2)^2 - 4*log(2) + 1) + 1/2*(2*x^2*log(x)^2 - 2 *x^2*log(x) + x^2)/(4*log(2)^2 - 4*log(2) + 1) + 1/2*(2*x^2*log(x) - x^2)/ (4*log(2)^2 - 4*log(2) + 1) + x/(4*log(2)^2 - 4*log(2) + 1)
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=x^{2} \log \left (x\right )^{2} + \frac {25 \, \log \left (-\log \left (2\right ) + \log \left (x\right )\right )^{4}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + x \] Input:
integrate((100*log(log(1/2*x))^3+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1 /2*x)*log(x)^2+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1/2*x)*log(x)+(4*x* log(2)^2-4*x*log(2)+x)*log(1/2*x))/(4*x*log(2)^2-4*x*log(2)+x)/log(1/2*x), x, algorithm="giac")
Output:
x^2*log(x)^2 + 25*log(-log(2) + log(x))^4/(4*log(2)^2 - 4*log(2) + 1) + x
Time = 7.64 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=\frac {\left (8\,{\ln \left (2\right )}^2-\ln \left (256\right )+2\right )\,x^2\,{\ln \left (x\right )}^2}{2\,\left (4\,{\ln \left (2\right )}^2-\ln \left (16\right )+1\right )}+x+\frac {{\ln \left (\ln \left (\frac {x}{2}\right )\right )}^4}{4\,\left (\frac {{\ln \left (2\right )}^2}{25}-\frac {\ln \left (16\right )}{100}+\frac {1}{100}\right )} \] Input:
int((100*log(log(x/2))^3 + log(x/2)*(x - 4*x*log(2) + 4*x*log(2)^2) + log( x/2)*log(x)^2*(8*x^2*log(2)^2 - 8*x^2*log(2) + 2*x^2) + log(x/2)*log(x)*(8 *x^2*log(2)^2 - 8*x^2*log(2) + 2*x^2))/(log(x/2)*(x - 4*x*log(2) + 4*x*log (2)^2)),x)
Output:
x + log(log(x/2))^4/(4*(log(2)^2/25 - log(16)/100 + 1/100)) + (x^2*log(x)^ 2*(8*log(2)^2 - log(256) + 2))/(2*(4*log(2)^2 - log(16) + 1))
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.59 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=\frac {25 \mathrm {log}\left (\mathrm {log}\left (\frac {x}{2}\right )\right )^{4}+4 \mathrm {log}\left (x \right )^{2} \mathrm {log}\left (2\right )^{2} x^{2}-4 \mathrm {log}\left (x \right )^{2} \mathrm {log}\left (2\right ) x^{2}+\mathrm {log}\left (x \right )^{2} x^{2}+4 \mathrm {log}\left (2\right )^{2} x -4 \,\mathrm {log}\left (2\right ) x +x}{4 \mathrm {log}\left (2\right )^{2}-4 \,\mathrm {log}\left (2\right )+1} \] Input:
int((100*log(log(1/2*x))^3+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1/2*x)* log(x)^2+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1/2*x)*log(x)+(4*x*log(2) ^2-4*x*log(2)+x)*log(1/2*x))/(4*x*log(2)^2-4*x*log(2)+x)/log(1/2*x),x)
Output:
(25*log(log(x/2))**4 + 4*log(x)**2*log(2)**2*x**2 - 4*log(x)**2*log(2)*x** 2 + log(x)**2*x**2 + 4*log(2)**2*x - 4*log(2)*x + x)/(4*log(2)**2 - 4*log( 2) + 1)